Find a function less than a fraction of itself

In summary, The conversation discusses the task of finding a function ##u(x)## that satisfies the inequality ##|u(x)-u(y)| \leq 0.5|x-y|## for all ##x,y \in I##, where ##I=(-1,1)##. The function should also not have the property ##u(x) = x## for any ##x\in I##. The conversation delves into possible solutions such as linear functions and sine and polynomial functions, but ultimately concludes that the function must be continuous and not cross the line ##y=x## in order to satisfy the given conditions.
  • #1
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Homework Statement
Give an example of a function ##u## such that ##u:I\to I=(−1,1)## such that ##|u(x)−u(y)|\leq 0.5|x−y|## for all ##x,y\in I##, and ##u(x) \neq x## for all ##x\in I##.
Relevant Equations
Nothing comes to mind
Well doesn't ##u(x) = 0.4 x## work? Seems too easy, but the phrasing at the end "for all ##x\in I##" makes me think since ##0.4x = x## only at ##x=0##, and not all of ##I##, that this is okay. But am I wrong?
 
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  • #2
##u(x) \neq x## for all ##x\in I## means, that none of the ##x## should have this property. Thus linear functions are ruled out.
 
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  • #3
fresh_42 said:
##u(x) \neq x## for all ##x\in I## means, that none of the ##x## should have this property. Thus linear functions are ruled out.
But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
 
  • #4
joshmccraney said:
But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].
 
  • #5
Mark44 said:
How about fractional powers? You want a function that is bounded above by y = (1/2)x on [0, 1] and bounded below by the same function on [-1, 0].
But at ##x=0## all fractional powers are zero. So it won't work, right?
 
  • #6
joshmccraney said:
But at ##x=0## all fractional powers are zero. So it won't work, right?
You could just make the function undefined at x = 0.
 
  • #7
Mark44 said:
You could just make the function undefined at x = 0.
But the function requires ##I\to I##.
 
  • #8
joshmccraney said:
But the function requires ##I\to I##.
I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement
and ##u(x) \neq x## for all ##x\in I##.
to mean that u(x) is not identically equal to x. I read this as different from ##\forall x \in (-1, 1), f(x) \ne x##; i.e., for each x in I, ##f(x) \ne x##.
 
  • #9
Mark44 said:
I don't think that precludes omitting 0 in how the function is defined.

However, if you don't like that, one can interpret the requirement to mean that u(x) is not identically equal to x. I read this as different from ##\forall x \in (-1, 1), f(x) \ne x##; i.e., for each x in I, ##f(x) \ne x##.
So is there even such a function that takes ##I\to I##, satisfying the inequality, where it does not cross the line ##y=x## anywhere in ##I##? I also read it that way at first, but it seemed way too easy.
 
  • #10
joshmccraney said:
So is there even such a function that takes ##I\to I##, satisfying the inequality, where it does not cross the line ##y=x## anywhere in ##I##? I also read it that way at first, but it seemed way too easy.

Can you show that ##u## must be continuous?

Can you draw such a function? Maybe forget formulas to begin with. What would it look like?
 
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  • #11
PS if ##u## must also be onto, then it is impossible. You might want to try proving this.

Hint: assume first that ##u(0) > 0## and reach a contradiction to the required inequality.
 
  • #12
As I read it, the function doesn't need to be surjective, but defined everywhere on ##I=(-1,1)## as the condition involves all points of ##I##. This means you have three tasks:
a) let ##u(x)## be a composite function, e.g. two linear ones with a 'jump' at ##x=0##
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval ##[-1,1]##
c) this leads to what is probably meant by the exercise: ##u(x)=x## has to be at the points ##x=\pm 1## for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin
 
  • #13
There is actually a simple solution. A simple family of solutions, in fact.
 
  • #14
What about the function ##u(x) = 2##? Then ##|u(x)-u(y)| = |2-2| = 0 \leq 0.5|x-y|##. Additionally, in ##I## we note that ##u(x) \neq x##.
 
  • #15
joshmccraney said:
What about the function ##u(x) = 2##? Then ##|u(x)-u(y)| = |2-2| = 0 \leq 0.5|x-y|##. Additionally, in ##I## we note that ##u(x) \neq x##.

##2 \notin I##
 
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  • #16
PeroK said:
##2 \notin I##
Ahhhhh bummer! Ok ok, let me think about what you posted in your previous comment.
 
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  • #17
fresh_42 said:
As I read it, the function doesn't need to be surjective, but defined everywhere on ##I=(-1,1)## as the condition involves all points of ##I##. This means you have three tasks:
a) let ##u(x)## be a composite function, e.g. two linear ones with a 'jump' at ##x=0##
b) prove what @PeroK said: there is no such function if it has to be surjective on the closed interval ##[-1,1]##
c) this leads to what is probably meant by the exercise: ##u(x)=x## has to be at the points ##x=\pm 1## for otherwise it is not possible, which leads back to @Mark44 's parabola - just lift it a bit above the origin
A composite linear function with a jump at ##x=0## seems impossible to force ##|u(x)-u(y)| \leq 0.5|x-y|## since the ##\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u##, yet ##\lim_{x\to 0} 0.5x = 0##. Am I missing something?
 
  • #18
joshmccraney said:
Given a composite linear function with a jump at ##x=0## seems impossible to force ##|u(x)-u(y)| \leq 0.5|x-y|## since the ##\lim_{x \to 0^-}u \neq \lim_{x \to 0^+}u##, yet ##\lim_{x\to 0} 0.5x = 0##. Am I missing something?

The function must be continuous. The only real constraint is that ##u## must not cross the line ##y=x##.

Why not keep ##u## above that line?

When you tried ##u(x) = 2##, you would have been closer with ##u(x) = 1##.
 
  • #19
PeroK said:
The function must be continuous. The only real constraint is that ##u## must not cross the line ##y=x##.

Why not keep ##u## above that line?

When you tried ##u(x) = 2##, you would have been closer with ##u(x) = 1##.
Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!
 
  • #20
joshmccraney said:
Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!

You don't need a V.
 
  • #21
joshmccraney said:
Yea, I was about to respond with that but ##1\notin I##.

If we agree the function must be continuous, then are you saying make like a ##V## but not cross ##y=x##? Seems like a good solution!

Okay. Go for a V. I keep looking at the OP to see whether I'm missing something!
 
  • #22
joshmccraney said:
But then ##\sin(x)## is also ruled out? And so is ##x^{2}##. Any ideas?
How about rescaling something that is increasing , with |f(-1)|=|f(1)|? Just rescale by the value at either endpoint.
 
  • #23
Wait, I totally spaced the function is ##y=0.5x## and not ##y=x##. In this case ,why not ##u(x) = 0.7##? Doesn't this work?
 
  • #24
joshmccraney said:
Wait, I totally spaced the function is ##y=0.5x## and not ##y=x##. In this case ,why not ##u(x) = 0.7##? Doesn't this work?

I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

And, indeed, any differentiable function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

The same for functions that lie below the line ##y = x## and intersect it at ##x = -1##.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) ##|u(x) - u(y)| \le 0.5 |x-y|##

This enforces continuity and means the function can't be too steep.

b) ##u(x) \ne x##

This means that the function must lie above or below the line ##y=x##.

For example, the straight line joining ##(-1, 0)## with ##(1, 1)## would do. It has a constant gradient of ##1/2## and intersects the line ##y=x## outside the open interval ##I##.
 
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  • #25
PeroK said:
I can't understand the issue here. It's ptobably time to wrap this up and move on. Any linear function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

And, indeed, any differentiable function for which ##u(1) = 1## and whose gradient is ##\le 1/2## would do.

The same for functions that lie below the line ##y = x## and intersect it at ##x = -1##.

Maybe there was something about the working of the problem that confuses everyone. Maybe it was too much focus on formulas rather than a focus on some simple geometric constraints.

The only two constraints were

a) ##|u(x) - u(y)| \le 0.5 |x-y|##

This enforces continuity and means the function can't be too steep.

b) ##u(x) \ne x##

This means that the function must lie above or below the line ##y=x##.

For example, the straight line joining ##(-1, 0)## with ##(1, 1)## would do. It has a constant gradient of ##1/2## and intersects the line ##y=x## outside the open interval ##I##.
Yea sorry, I was really spacing out. But thanks! Kinda too late for the HW anyways, buuuuuuut I appreciate you sticking with it!

So if the interval ##I = \mathbb R## and we require the inequality be strictly less than, then would there be any such ##u(x)##?
 
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  • #26
joshmccraney said:
Yea sorry, I was really spacing out. But thanks! Kinda too late for the HW anyways, buuuuuuut I appreciate you sticking with it!

So if the interval ##I = \mathbb R## and we require the inequality be strictly less than, then would there be any such ##u(x)##?

It's fairly easy to see that there is no function. Just look at ##u(0)## and apply the MVT using the inequality given.
 
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  • #27
PeroK said:
It's fairly easy to see that there is no function. Just look at ##u(0)## and apply the MVT using the inequality given.
Got it, thanks. Didn't think about MVT, but it makes sense.
 
  • #28
joshmccraney said:
Got it, thanks. Didn't think about MVT, but it makes sense.
Although, in fact I meant the intermediate value theorem.
 

1. What does it mean to find a function less than a fraction of itself?

When we say "find a function less than a fraction of itself," we are referring to finding a mathematical function that is smaller in value than a fraction of itself. This can also be interpreted as finding a function with a decreasing rate of change.

2. How do you find a function less than a fraction of itself?

To find a function less than a fraction of itself, you can use various methods such as integration, differentiation, or algebraic manipulation. The specific method will depend on the given function and the fraction it needs to be less than.

3. Why is finding a function less than a fraction of itself important?

Finding a function less than a fraction of itself is important in many areas of mathematics and science, including optimization, economics, and physics. It allows us to model and analyze real-world situations where values are decreasing over time or distance.

4. Can you provide an example of finding a function less than a fraction of itself?

One example of finding a function less than a fraction of itself is modeling the decay of a radioactive substance. The function representing the amount of remaining substance will be less than a fraction of itself as time passes, and we can use this to predict the rate of decay.

5. Are there any limitations to finding a function less than a fraction of itself?

Yes, there are limitations to finding a function less than a fraction of itself. In some cases, the function may not exist or may not be able to be expressed in a closed form. Additionally, the accuracy of the function may be limited by the precision of the given data or the assumptions made in the modeling process.

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